Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.
Note: m and n will be at most 100.
简单的动态规划问题
class Solution {
public:
int dp[][];
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
memset(dp,,sizeof(dp));
dp[][]=obstacleGrid[][] == ? :;
int n = obstacleGrid.size(), m = obstacleGrid[].size();
for(int i = ; i < n; ++ i )
dp[i][] = obstacleGrid[i][] == ? : dp[i-][];
for(int i = ; i < m; ++ i )
dp[][i] = obstacleGrid[][i] == ? : dp[][i-];
for(int i = ; i < n ; ++ i){
for(int j = ; j < m ; ++ j){
dp[i][j] = obstacleGrid[i][j] == ? : dp[i-][j] + dp[i][j-];
}
}
return dp[n-][m-];
}
};